Question

Assume that the differences are normally distributed. Complete parts ​(a) through ​(d) below.

Observation

1

2

3

4

5

6

7

8

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Upper X Subscript iXi

46.746.7

47.747.7

45.645.6

50.250.2

48.448.4

50.850.8

47.847.8

48.648.6

Upper Y Subscript iYi

50.150.1

48.448.4

47.347.3

54.554.5

47.947.9

50.950.9

49.649.6

50.350.3

​(a) Determine

d Subscript i Baseline equals Upper X Subscript i Baseline minus Upper Y Subscript idi=Xi−Yi

for each pair of data.

Observation	1	2	3	4	5	6	7	8
di	negative 3.4 −3.4	negative 0.7 −0.7	negative 1.7 −1.7	negative 4.3 −4.3	0.5 0.5	negative 0.1 −0.1	negative 1.8 −1.8	negative 1.7 −1.7

​(Type integers or​ decimals.)

​(b) Compute

d overbard

and

s Subscript dsd.

d overbardequals=negative 1.650 −1.650

​(Round to three decimal places as​ needed.)

s Subscript dsdequals=1.605 1.605

​(Round to three decimal places as​ needed.)​(c) Test if

mu Subscript dμdless than<0

at the

alphaαequals=0.05

level of significance.

What are the correct null and alternative​ hypotheses?

A.

Upper H 0H0​:

mu Subscript dμdless than<0

Upper H 1H1​:

mu Subscript dμdequals=0

B.

Upper H 0H0​:

mu Subscript dμdgreater than>0

Upper H 1H1​:

mu Subscript dμdless than<0

C.

Upper H 0H0​:

mu Subscript dμdless than<0

Upper H 1H1​:

mu Subscript dμdgreater than>0

D.

Upper H 0H0​:

mu Subscript dμdequals=0

Upper H 1H1​:

mu Subscript dμdless than<0

Answer 1

(a)

is given by:

Observation	1	2	3	4	5	6	7	8
di	- 3.4	-0.7	- 1.7	- 4.3	0.5	- 0.1	- 1.8	- 1.7

(b)

(i)

= - 13.2/8 = - 1.650

(ii)

(c)

Correct option:

D.   H0:

       H1:

(d)

SE = s_d/

= 1.605/ = 0.5675

Test statistic is:

t = - 1.650/0.5675 = - 2.9077

= 0.05

ndf = 8 - 1 = 7

One Tail Left Side Test

From Table, critical value of t = - 1.8946

Since the calculated value of t = - 2.9077 is less than critical value of t = - 1.8946, the difference is significant. Reject null hypothesis.

Conclusion:

The data support the claim that < 0 at the = 0.05 level of significance.